Page|85
(2, 0)
(0,β3)
Characteristics of Polynomial Functions2.1
POLYNOMIAL FUNCTIONS
2.1 Characteristics of Polynomial Functions p. 85
2.2 The Remainder Theorem p. 97
2.3 The Factor Theorem p. 109
REVIEW PRACTICE SECTION p. 131
2.4 Analyzing Graphs of Polynomial Functions p. 121
Determine an equation for each of the following functions:
Youβve already studied Polynomial functions! Remember?
In Math 10C you studied Linear Functions And in Math 20-1 you studied Quadratic Functions
ππ = ππππ+ ππ
We saw how equations of linear functions can be written in the form
ππ = ππ(ππβ ππ)ππ+ππ
β¦ And equations of quadratic functions can be written
Where ππ is the slope of the line, and ππ is the ππ-intercept
Where ππ is the vertical stretch, and the coordinates of the vertex are (ππ,ππ).
(β,ππ)
(ππ, 0) (ππ, 0)
ππ = ππ(ππβππ)(ππβ ππ)Note that the quadratic functions can also be written in the form
Where ππ, ππ are ππ-intercepts
ππ = ππ(ππβ ππ)Note that the linear functions can also be written in the form
Where ππ is the ππ-intercept
(0,ππ)(ππ, 0)
1
(1, 0)
(β1,β2)
(β3, 0)
2
Hint:The vertical stretch here, ππ, is ππ/ππ
These are degree 1 Polynomial Functions These are degree 2 Polynomial Functions
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2.1 Characteristics of Polynomial Functions
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Identifying Polynomial FunctionsClass Example 2.11
(a) π¦π¦ = 4π₯π₯5 β 3π₯π₯2
Identify which of the following are polynomial functions. For each that is a polynomial function; state the degree and leading coefficient:
(b) π¦π¦ = 2 π₯π₯ + 5π₯π₯ (c) π¦π¦ = 3 π₯π₯ + 7 (d) π¦π¦ = β2π₯π₯3 + 5π₯π₯β1
Worked Example
Identify which of the following are polynomial functions. For each that is a polynomial function; state the degree and leading coefficient:
(a) π¦π¦ = β6π₯π₯2 +3π₯π₯β 8 (b) π¦π¦ = 3π₯π₯4 β 2π₯π₯5 β 3π₯π₯2 β 1 (c) π¦π¦ = 5π₯π₯2 β 33 π₯π₯ + 1
Solution: (a) The middle term can be written 3π₯π₯β1, which is NOT POLYNOMIAL as exponent of π₯π₯ is not a whole number
(b) All exponents are whole numbers, and all coefficients are real numbers. Hello, you POLYNOMIAL FUNCTION.Degree is 5 (degree of entire poly function is that of highest degree term). Leading coefficient is βππ. (the terms are not in descending order of degree β the 2nd term should be re-arranged to the βfrontβ!)
(c) The middle term can be written 3π₯π₯1/3, which is NOT POLYNOMIAL as the exponent is not a whole number
A polynomial function, such as a linear function, quadratic, or cubic, involves only non-negative integer powers of π₯π₯.
Any polynomial function can be written in the formππ π₯π₯ = πππππ₯π₯ππ + ππππβ1π₯π₯ππβ1 + ππππβ2π₯π₯ππβ2 + β¦ + ππ2π₯π₯2 + ππ1π₯π₯1 + ππ0
Where: ππ is a whole number, representing the degree of the functionππππ to ππ0 are real numbers, representing the coefficients, with ππππdesignated the leading coefficient (coeff. of the highest degree term)
This general formula may look complicated, but a few polynomial function examples should show its simplicity:
π¦π¦ = 8 This is degree 0 (constant function)
π¦π¦ = β5π₯π₯ + 4 This is degree 1 (linear), with a leading coefficient of β5
ππ(π₯π₯) = 2π₯π₯2 β 4π₯π₯ + 3 This is degree 2 (quadratic), with a leading coefficient of 2
π¦π¦ = π₯π₯3 +12π₯π₯2 β 7π₯π₯ β 1 This is degree 3 (quadratic), with a leading coefficient of 1
ππ π₯π₯ = 3π₯π₯4 β 2π₯π₯3 + 5π₯π₯ β 8 This is degree 4 (quartic), with a leading coefficient of 3
ππ π₯π₯ = β2π₯π₯5 + π₯π₯4 β 3π₯π₯2 β 6 This is degree 5 (quintic), with a leading coefficient of β2
Polynomial functions can be of any whole number degree ππ β but for this course weβll only deal with functions where ππ β€ 5.
And while the coefficients can be any real number β weβll mainly stick with integer coefficients.
These examples are all written in descending order of degree, where terms are arranged starting with the highest degree term, starting with the leading coefficient. (The coefficient of the highest degree term)
Defining Polynomial Functions
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Unit 2 β Polynomial Functions
For both terms, exponents of π₯π₯ are whole numbers. POLYNOMIAL FUNCTION , degree 5 Leading Coeff. 4(a)
Answers from previous page
2.11The first term could be written as 2π₯π₯1/2, which is NOT POLYNOMIAL as the exponent of π₯π₯ is not a whole number. (b)
Thatβs a POLYNOMIAL FUNCTION (Donβt be thrown off by the irrational 3 coefficient β thatβs allowed!) Polynomial functions must have whole # variable exponents, coefficients can be any real #. degree 1 Leading Coeff. ππ
(c)
The last term is NOT POLYNOMIAL as the exponent of π₯π₯ is not a whole number. (d)
A polynomial function can be of any whole number degree, including zero!The graph on the right is of the constant function ππ π₯π₯ = 3, which is a polynomial function of degree zero. Letβs now acquaint ourselves with some examples of polynomial functions, of degree 1 through 5:
ππ π₯π₯ = 2π₯π₯ β 3
Degree 1 Linear
Domain: {π₯π₯ β β}Range: {π¦π¦ β β}
End Behavior: starts negative in quad III, ends positive in quad I# of intercepts: ππ
ππ π₯π₯ = βπ₯π₯2 + 2π₯π₯ + 3
Degree 2 Quadratic
Domain: {π₯π₯ β β}Range: {π¦π¦ β€ 4, π¦π¦ β β}
End Behavior: starts negative in quad III, ends negative in quad IV# of intercepts: ππ
ππ π₯π₯ = π₯π₯4 β 9π₯π₯2 β 4π₯π₯ + 12
Degree 4 Quartic
Domain: {π₯π₯ β β}Range: {π¦π¦ β₯ β16.9, π¦π¦ β β}
End Behavior: starts positive in quad II, ends positive in quad I# of intercepts: ππ
ππ π₯π₯ = π₯π₯3 β 3π₯π₯2 β π₯π₯ + 3
Degree 3 Cubic
Domain: {π₯π₯ β β}Range: {π¦π¦ β β}
End Behavior: starts negative in quad III, ends positive in quad I# of intercepts: ππ
The functions on the left are odd degree β and the graphs start and end in the opposite direction. For example, the degree 5 function starts positive and ends negative.
Odd functions have no max or min point, must have at least one π₯π₯-intercept, and have a range {ππ β β}.
The functions above are even degree. As such, the graphs start and end in the same direction. For example, the degree 4 function starts positive and ends positive.
Even functions have either a maximum or minimum point, and the range is restricted accordingly.
If the sign of the leading coefficient is positive (the degree 1, 3, and 4 examples above), the graph βends positiveβ, or heading upward, in quad I. And If the degree is even, the graph will have a minimum point.
I wish my lead coeff. wasnβt so negative
Ends positive
If the sign is negative (see the degree 2 and 5 examples), the graph βend negativeβ, or downward, in quad IV. And if its even degree (as with example 2), the graph will have a maximum point.
Classifying Polynomial Functions by Degreeππ π₯π₯ = 3
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β π₯π₯ = βπ₯π₯5 + 4π₯π₯4 + π₯π₯3 β 16π₯π₯2 + 12π₯π₯
Degree 5 Quintic
Domain: {π₯π₯ β β}Range: {π¦π¦ β β}
End Behavior: starts positive in quad II, ends negative in quad IV# of intercepts: ππ
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2.1 Characteristics of Polynomial Functions
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There is a relationship between the degree of a polynomial function and the number of ππ-intercepts on the graph.
For a polynomial function of degree ππ; the maximum number of π₯π₯-intercepts is ππ.
For example, this degree 4function has no π₯π₯-intercepts
For example, consider a degree 5 functionSince it starts / ends in the opposite direction (in this case starts positive in quad II, ands negative in quad IV)β¦there must be at least one ππ-intercept
For even degree functions, there can be ππ to ππ π₯π₯-intercepts.
For odd degree functions, there can be ππ to ππ π₯π₯-intercepts.
Now, a degree 4 function can also have 1 ππ-intercept β¦ or 2 ππ-intercepts β¦ or 3 ππ-intercepts To a maximum of 4
Note (unlike the functions above), this functionmust have a negative leading coefficient. (ends down)
And we should know how to spot a Polynomial Function Graph!
On the previous page we saw the relationship between the degree of a polynomial function and certain characteristics of the graph. You might next ask βhow can we immediately tell that a graph is of a polynomial function, and not some other function we study in Math 30?
And once again β great question! I respect your inquisitive nature. Letβs dive into that, with a couple of key distinguishing points:
Have no start or end points, like, for example, radical function graphs.
Polynomial Function
Radical Function
(Domain is Restricted)
Graph starts at this point
Have no vertical asymptotes or any other type of discontinuity, as with rational function graphs.
Graph can be drawn without lifting your pencil.
The first key point is that all polynomial functions have a domain {π₯π₯ β β}. That means graphs of polynomial functions:
ππ π₯π₯ = 0.5 π₯π₯ + 3 β 2 ππ π₯π₯ =1
(π₯π₯ + 2)(π₯π₯ β 3)
The second point is polynomial function graphs have no horizontal asymptotes (like exponential functions) and there is no periodic pattern (as with some trig graphs).
So graphs will always both start and end in either an upward or downward position.
Rational Function
Exponential Function
β π₯π₯ = (1.5)π₯π₯β3
ENDS upward (pos lead coeff.)
STARTS pointing downward
For example, this graphβ¦.
Graph has vertical asymptotes (again, domain is restricted)
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Unit 2 β Polynomial Functions
Characteristics of Polynomial FunctionsClass Example 2.12
(a) ππ = ππππ β ππππππ β ππππ + ππ
For each of the following polynomial functions, without using your graphing calculator, state:
(b) ππ = ππππ + ππππ β ππππππ β ππππππππ β ππππ
i - The start and end behavior of the graph ii - The number of possible π₯π₯-interceptsiii - Whether or not the graph will have a minimum or maximum pointiv - The domain of the function and the π¦π¦-intercept
Use your graphing calculator to determine:v - The π₯π₯-intercepts of the graph vi - The range of the function
Worked Example
For the polynomial function ππ π₯π₯ = βπ₯π₯4 + 3π₯π₯3 + 7π₯π₯2 β 15π₯π₯ β 18 ;
Sol.:
Without using your graphing calculator, state:i - The start and end behavior of the graph ii - The number of possible π₯π₯-interceptsiii - Whether or not the graph will have a minimum or maximum pointiv - The domain of the function and the π¦π¦-intercept
Use your graphing calculator to determine:v - The π₯π₯-intercepts of the graph vi - The range of the function
i -
The degree of the function is 4; since itβs even, the graph will start and end in the same direction. And since the leading coefficient is negative (that is, β-1β), the graph will end negative / heading downward. Soβ¦. The graph starts negative in quadrant III, and ends negative in quadrant IV.
ii - Degree 4 (even), so there can be between 0 and 4 ππ-intercepts. iii - Even degree, graph starts / ends in the same direction. Negative leading coefficient, so which means the
graph ends negative. Therefore the graph will have a maximum point, which can be found graphically.iv - All polynomial functions have domain {ππ β β}. The π¦π¦-intercept is the same as the constant value, so (ππ,βππππ)
So, ππ-intercepts are (βππ,ππ), (βππ,ππ), and (ππ,ππ)
vi -For the range, find the MAXIMUM, which is also in the CALC menu.
v - π₯π₯-intercepts are the same as the zeros of the function. The zero function is in CALC menu, found be entering
Find the zeros one at a time...
ππ = βππ
ππ = βππ
Note: sometimes the calc adds decimals. Here, the actual value is just ππ.
The range is: {ππ β€ ππ.ππππ,ππ β β}Note that the maximum is provided as an approximate value, to the nearest hundredth.
i - Start / end
ii - # of π₯π₯-ints
iii - Max or min?
iv - Domain:π¦π¦-intercept:
v - Coords of π₯π₯-ints:
vi - Range:
i - Start / end
ii - # of π₯π₯-ints
iii - Max or min?
iv - Domain:π¦π¦-intercept:
v - Coords of π₯π₯-ints:
vi - Range:
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2.1 Characteristics of Polynomial Functions
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More Characteristics of Polynomial FunctionsClass Example 2.13
For the function ππ ππ = βππππ β ππππππ + ππππππ + ππππππβ ππππ, without using your graphing calculator, state:
i - The start and end behavior of the graph
ii - The number of possible π₯π₯-intercepts
iii - Whether or not the graph will have a minimum or maximum point
iv - The domain of the function and the π¦π¦-intercept
Then, use your graphing calculator to determine:v - The π₯π₯-intercepts of the graph
vi - The range of the function nearest hundredth
Label any intercepts and max / min pointsLastly, sketch the graph on the grid below.
i β Odd degree, Positive Leading Coefficient β So starts negative in quadrant III and ends positive in quad I.(a)2.12ii β Degree 3 (odd), so between 1 and 3 π₯π₯-intercepts. iii β Odd degree so no max / miniv β Domain is {ππ β β} π¦π¦-intercept is: π¦π¦ = 0 3 β 2 0 2 β β6 0 + 6 = 6 So cords: (ππ,ππ)v β π₯π₯-intercepts (use ZERO on calc): βππ,ππ , (ππ,ππ), and (ππ,ππ). vi β Range: {ππ β β}.
i β Odd degree, Positive Leading Coefficient β So starts negative in quadrant III and ends positive in quad I.(b)ii β Degree 5 (odd), so between 1 and 5 π₯π₯-intercepts. iii β Odd degree so no max / miniv β Domain is {ππ β β} π¦π¦-intercept is: π¦π¦ = (0)5+(0)4β7 0 3 β 13 0 2 β 6(0) = ππ So cords: (ππ,ππ)v β π₯π₯-intercepts: βππ,ππ , (βππ,ππ), (ππ,ππ), and (ππ,ππ). vi β Range: {ππ β β}.
Answers from previous page
Identifying Polynomial FunctionsClass Example 2.14
For each of the polynomial functions listed below, indicate the graph number that matches.
(a) π¦π¦ = π₯π₯4 β 2π₯π₯3 β 7π₯π₯2 + 8π₯π₯ + 12
(b) π¦π¦ = βπ₯π₯5 + 11π₯π₯3 + 6π₯π₯2 β 28π₯π₯ β 24
(c) π¦π¦ = π₯π₯5 β π₯π₯4 β 9π₯π₯3 + 13π₯π₯2 + 8π₯π₯ β 12
(d) π¦π¦ = π₯π₯4 β 4π₯π₯3 β π₯π₯2 + 16π₯π₯ β 12
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Unit 2 β Polynomial Functions
Applications of Polynomial Functions
In Math 20, we saw how a certain type of polynomial function, the quadratic (degree 2) function, has applications in parabolic motion and finance. (To name just two!)
Remember finding the maximum height of a ball?
Above β classic math 20 question, do you remember how to find the max height?
i β Starts neg. in quad. III, ends neg. IV. ii β Can be 0 to 4 π₯π₯-ints (a)2.13iii β Even degree, neg lead coeff., so graph has MAXiv β Domain is {ππ β β} π¦π¦-intercept is: (ππ,βππππ)v β π₯π₯-ints: βππ,ππ , (ππ,ππ), (ππ,ππ). vi β Range: {ππ β€ ππ.ππππ,ππ β β}.
For RANGE, find the MAX:
(a) (b) (c) (d) 2.14
Answers from previous page
A box is with no lid is made by cutting four squares (each with a side length βπ₯π₯β from each corner of a 24 cm by 12 cm rectangular piece of cardboard.
(a) Determine a function that models the volume of the box.
(b) Use technology to graph the function, and sketch below. Label each axis, provide a scale, and indicate any intercepts or max / min points.Use your graphing calculator to obtain theseβ¦ youβll need to βtrial-and-errorβ a suitable viewing window, indicate in your sketch below.
(c) State the domain of the function, with respect to the βreal-worldβ constraints of the problem.
(e) State the maximum volume of the box, (Round to the nearest ππππ3)
Constructing and Analyzing a Polynomial EquationClass Example 2.15
First β match the window to whatβs given. Graph ππππ = βππππππππ + ππππππππThen β from the CALC menu, select #4, βMAXIMUMβ
Next β For βleft boundβ, hit anywhere to the left of the max. Do the same for βright boundβ, anywhere to the right.
The MAX should be between these arrows.
Finaly β Hit for βGuessβ. The max value is the π¦π¦-coord.
So, the maximum height of the ball is 156.24 feet, after 3.1 seconds.
(d) State the value of βπ₯π₯β that gives the maximum volume. (Round to the nearest hundredth)
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2.1 Characteristics of Polynomial Functions
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A box with a lid can be created by removing two congruent squares from one end of a rectangular 8.5 inch by 11 inch piece of cardboard. The congruent rectangles removed from the other end as shown. (The shaded rectangles represent the waste, or removed portions that will not be used in the box)
(a) In the diagram below there are two congruent rectangles; one that will form the base of the box, and one that will be the top. Complete the diagram by providing the missing dimensions (indicated with / ) for the base and top.
Constructing and Analyzing a Polynomial EquationClass Example 2.16
ππππ
ππππ inches
ππ.ππinches
ππ.ππ inches ππ.ππ inches
(b) Determine a function that models the volume of the box.
(c) Use technology to graph the function, and sketch below. Label each axis, provide a scale, and indicate any intercepts or max / min points.Use your graphing calculator to obtain theseβ¦ youβll need to βtrial-and-errorβ a suitable viewing window, indicate in your sketch below.
(d) State the domain of the function, with respect to the βreal-worldβ constraints of the problem.
(f) State the maximum volume of the box, (Round to the nearest thousandth)
(e) State the value of βπ₯π₯β that gives the maximum volume. (Round to the nearest thousandth)
(g) State the dimensions that yield the maximum volume. (Round to the nearest thousandth)
ππππ
ππ = π₯π₯(12 β 2π₯π₯)(24 β 2π₯π₯)(a)2.15
Answers from previous page
Note that the part of the graph to the right of π₯π₯ = 6is βirrelevantβ, as the y-coord (Volume) is negative.
Only considerthis portion of graph
(c) Domain is [ππ,ππ](d) Max when ππ β ππ.ππππ cm x-coord of MAX point
(e) Max Volume: β ππππππ ππππππ y-coord of MAX
Remember units!
(b)(ππ.ππππ,ππππππ.ππππ)
(ππ,ππ)(ππ,ππ)
Your sketch should only be shown within the domain, [0,6]. Show dots () on the graph at the domain start & end points, just like here.
Label the coords of the intercepts and max point. Provide a scale, and label what each axis represents.
Volu
me
Length cut out
ππππ β ππππππππ β ππππ
Note: Do not expand function!
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Unit 2 β Polynomial Functions
2.1 Practice Questions
1. Indicate which of the following functions are polynomial functions:
(a) π¦π¦ = 3π₯π₯5 β 3π₯π₯3 + 2π₯π₯2 + 11π₯π₯ + 6
(d) π¦π¦ = 4π₯π₯4 β 2π₯π₯2 + 5π₯π₯β1 β 1
(b) π¦π¦ = 3π₯π₯3 β 5π₯π₯0.5 + 2
(e) π¦π¦ = 3π₯π₯3 β 5 π₯π₯
(c) π¦π¦ = 5
(f) π¦π¦ = π₯π₯2 + 5π₯π₯ + 2
2. Indicate which of the following graphs are likely those of polynomial functions:(a) (b) (c) (d)
(e) (f) (g) (h)
3. For each of the following polynomial functions, state each of the indicated characteristics. Try as many as you can without graphing.
iii - Start / end behavior
iv - Possible # of π₯π₯-intercepts
v - Whether there is a max or min
vi -π¦π¦-intercept
i - Lead Coefficient
ii - Degree
(a) ππ(π₯π₯) = π₯π₯3 + 8π₯π₯2 + 11π₯π₯ β 20 (b) π¦π¦ = 5 β π₯π₯4
(c) π¦π¦ = β2π₯π₯4 β 6π₯π₯3 + 14π₯π₯2 + 30π₯π₯ β 36 (d) π¦π¦ = β2 π₯π₯ + 3 2(π₯π₯ β 2)2(π₯π₯ β 1)
i -
ii -
iii -
iv -
v -
vi -
(a) i -
ii -
iii -
iv -
v -
vi -
(b) i -
ii -
iii -
iv -
v -
vi -
(c) i -
ii -
iii -
iv -
v -
vi -
(d)
π½π½ = ππ(ππ.ππ β ππππ)(ππ.ππ β ππ)
(a)2.16
Answers from previous page
(d) Domain is [ππ,ππ.ππππ](e) Max when ππ β ππ.ππππππ in x-coord of MAX point
(f) Max Volume: β ππππ.ππππππ ππππππ y-coord of MAX
Remember units!
(ππ.ππππππ,ππππ.ππππππ)
(ππ.ππππ,ππ)
Label your graph! And only show the relevant portion / within domain(c)
(g) ππ.ππππππ Γ ππ.ππππ Γ ππ.ππππππ inches (b)
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4. For each of the following graphs, determine the indicated characteristics of the related function.
i -
ii -
iii -
iv -
v -
(a)
iii - # of π₯π₯-intercepts
iv - Range
v - Constant term in function equation
i - Is the degree even or odd?
ii - Is the leading coefficient pos (+) or neg (-)
i -
ii -
iii -
iv -
v -
(b) i -
ii -
iii -
iv -
v -
(c) i -
ii -
iii -
iv -
v -
(d)
(a)(b) (c)(a)
5. For each of the following functions, use technology to determine each of the indicated characteristics. Note that using technology (graphing on your calc) is not required for each characteristic each time! For example, see if you can spot the π₯π₯-intercepts of (c) without graphing. (And degree and π¦π¦-ints can always be found without graphing)Also note: To get best results graphing on your calculator β you must practice setting your window! For most of these you can use an ππ-min of βππ and an ππ-max of ππ. However for the ππmin and max β¦. use trial and error! (Youβll want to see any relative max / min points, so ensure your window is βlarge enoughβ)
iii - The coordinates of π¦π¦-intercept
iv - The Range
i - The degree
ii - The coordinates of any π₯π₯-intercepts
(a) ππ(π₯π₯) = π₯π₯3 + 8π₯π₯2 + 11π₯π₯ β 20 (b) π¦π¦ = π₯π₯4 β 3π₯π₯3 β 12π₯π₯2 + 52π₯π₯ β 48
(c) π¦π¦ = β π₯π₯ + 3 2(π₯π₯ β 1)(π₯π₯ β 3) (d) π¦π¦ = β2π₯π₯2 + 2π₯π₯ + 24
i -
ii -
iii -
iv -
(a) i -
ii -
iii -
iv -
(b) i -
ii -
iii -
iv -
(c) i -
ii -
iii -
iv -
(d)
Note: Where applicable, round to the nearest hundredth.
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Answers to Practice Questions on the previous page
Polynomial functions are: (a), (c), (e)1. Polynomial functions are: (a), (d), and (f)2.
3. (a) i ππ ii 3 iii Starts neg in quad III, ends pos in quad I iv 1 to 3 v No max or min vi (ππ,βππππ)(b) i βππ ii 4 iii Starts neg in quad III, ends neg in quad IV iv 0 to 4* v Graph has a max vi (ππ,ππ)(c) i βππ ii 4 iii Starts neg in quad III, ends neg in quad IV iv 0 to 4 v Graph has a max vi (ππ,βππππ)(d) i βππ ii 5* iii Starts pos in quad II, ends neg in quad IV iv 3* v No max or min vi (ππ,ππππ)
see note 1
see note 2see note 3
Note 1: We can visualize this, as the graph of π¦π¦ = π₯π₯4 is similar to π¦π¦ = π₯π₯2, so visualize a βparabolaβ opening down and shifted5 units up. So we know, without graphing, that there will be TWO π₯π₯-intercepts!
Note 2: For functions in factored form, the degree of the entire function is the sum of all exponents, so: 2 + 2 + 1 = πππ¦π¦ = β2 π₯π₯ + 3 ππ(π₯π₯ β 2)ππ(π₯π₯ β 1)
Note 3: Each factor corresponds to one π₯π₯-intercept, so we know with certainty there are 3. Thereβs an invisible β1β here!
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Unit 2 β Polynomial Functions
Answers to Practice Questions on the previous page
4. (a) i ODD ii NEGATIVE iii 4 π₯π₯-intercepts iv {ππ β β} v Constant term: βππππ (represented by π¦π¦-intercept)(b) i EVEN ii POSITIVE iii 3 π₯π₯-intercepts iv {ππ β₯ βππππ.ππππ,ππ β β} v Constant term: βππ(c) i ODD ii POSITIVE iii 2 π₯π₯-intercepts iv {ππ β β} v Constant term: βππππ(d) i EVEN ii NEGATIVE iii 3 π₯π₯-intercepts iv {ππ β€ ππππ.ππππ,ππ β β} v Constant term: ππ
(a) i ππ ii βππ,ππ , (βππ,ππ) and (ππ,ππ) iii (ππ,βππππ) iv {ππ β β}(b) i ππ ii βππ,ππ , (ππ,ππ) and (ππ,ππ) iii (ππ,βππππ) iv {ππ β₯ βππππππ.ππππ,ππ β β}(c) i ππ ii βππ,ππ , (ππ,ππ) and (ππ,ππ) iii (ππ,βππππ) iv {ππ β€ ππππ.ππππ,ππ β β}
Note: Each factor provides an π₯π₯-intercept(d) i ππ ii βππ,ππ and (ππ,ππ) iii (ππ,ππππ) iv {ππ β€ ππππ.ππ,ππ β β}
5.
6. Without graphing (use your reasoning abilities!), match each of the following functions with its graph.
(a) π¦π¦ = βπ₯π₯5 + 12π₯π₯3 + 2π₯π₯2 β 27π₯π₯ β 18
π¦π¦ = βπ₯π₯4 + π₯π₯3 + 11π₯π₯2 β 9π₯π₯ β 18(b)
π¦π¦ = π₯π₯5 β 2π₯π₯4 β 10π₯π₯3 + 20π₯π₯2 + 9π₯π₯ β 18(c)
π¦π¦ = βπ₯π₯4 + π₯π₯3 + 7π₯π₯2 β 13π₯π₯ + 6(d)
7. A package may be sent through a particular mail service only if it conforms to specific dimensions. To qualify, the sum of its height plus the perimeter of its base must be no more than 72 inches. Also for our design, the base of the box (shaded in the diagram below) has a length equal to double the width.
base
ππ =
2π₯π₯π₯π₯
(a) In the blank on the left, state an expression for the height (ππ) of the box.Need a hint? See the bottom of the next page.
(b) Determine a function that represents the Volume of the box.
(c) Use technology to graph the function obtained in (b) with a suitable viewing window. Provide your sketch below, labeling any max/mins and intercepts. Also fully label the axis, what each axis represents, and a suitable scale.
(d) Provide a domain and range for your function obtained in (b), with respect to the βreal worldβ constraints of the problem.
Domain: Range:
(e) State the maximum volume of the box that can be sent.
(f) State the dimensions for the box that provides the maximum volume.
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2.1 Characteristics of Polynomial Functions
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π½π½ = (ππππ)(ππ)(ππππ β ππππ)
(a)7.
Answers from previous and this page
(a) (b) (c) (d)
The perimeter of the base is: 2π₯π₯ + 2π₯π₯ + π₯π₯ + π₯π₯ = ππππ. As we wish for the largest volume box, weβll use all 72 inches (sum of perimeter and height) available. So β + 6π₯π₯ = 72, and ππ = ππππ β ππππ.
HINT for #7(a):
(d) Domain is [ππ,ππ.ππππ]
Max when ππ = ππ inches (f) Max Volume: = ππππππππ ππππππ
(g) ππππ length Γ ππ width Γ ππππ height inches
(ππ,ππππππππ)
(ππππ,ππ)(ππ,ππ)
Volu
me
width of box
Range is [ππ,ππππππππ]
Graph ππππ = in your calculator. Trial-and-error to get best window. Sketch should only show graph within domain. (between 0 and 6)
6.
ππ = ππππ β ππππ(c)
(b)
8. An open box is to be made by cutting out squares from the corners of an 8 inch by 15 inch rectangular sheet of cardboard and folding up the sides.
(a) On diagram 1 on the right, provide expressions that represent the length and width of the finished box.
(c) Use technology to graph the function, and sketch below. Label each axis, provide a scale, and indicate any intercepts or max / min points. Use your graphing calculator, provide a sketch below.
(d) State the domain and range of the function, with respect to the βreal-worldβ constraints.
(f) State the maximum volume of the box, (Round to the nearest ππππ3)
(e) State the value of βπ₯π₯β that gives the maximum volume. (Round to the nearest hundredth)
(b) Determine a function that models the volume of the box.
Diagram 1 Diagram 2
(g) Provide the dimensions that yield the box of maximum volume, (Round to the nearest hundredth)
π½π½ = (ππππ β ππππ)(ππ β ππππ)(ππ)
(a)8.(d) Domain is [ππ,ππ]
(e) Max when ππ = ππ.ππππ inches
(f) Max Volume: = ππππ.ππππ ππππππ
(g) ππππ.ππππ length Γ ππ.ππππ width Γ ππ.ππππheight inches
Range is [ππ,ππππ.ππππ]
(ππ.ππππ,ππππ.ππππ)
(ππ,ππ)(ππ,ππ)
Volu
me
Height
(b)
(c)
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